Consider the balanced reaction: Zn(s)+ 2 HCl(aq)→ ZnCl₂(aq)+ H₂(g). What volume of H₂(g)at STP can be generated when 134 g of zinc reacts?

Question

Quizplus · Accepted Answer

First, calculate the moles of Zn using its molar mass (65.38 g/mol). $ \frac{134 g}{65.38 g/mol} = 2.05 $ moles of Zn. The balanced equation shows that 1 mole of Zn produces 1 mole of $ H_2 $ gas. Therefore, 2.05 moles of Zn will produce 2.05 moles of $ H_2 $ gas. At STP, 1 mole of any gas occupies 22.4 L, so $ 2.05 $ moles of $ H_2 $ will occupy $ 2.05 \times 22.4 L = 45.9 L $.

Consider the Balanced Reaction: Zn(s)+ 2 HCl(aq)→ ZnCl2(aq)+ H2(g)

Consider the Balanced Reaction: Zn(s)+ 2 HCl(aq)→ ZnCl₂(aq)+ H₂(g)